B1.1 Carbohydrates and lipids

1. It forms up to four covalent bonds, single or double, with carbon or other non-metal atoms — A carbon atom has four outer electrons, so it can form up to four covalent bonds, as four single bonds or a combination of single and double bonds, with other carbon atoms or with atoms of other non-metallic elements. Repeating this gives chains, branches and rings.
2. It forms any number of covalent bonds, which is why so many carbon compounds exist — A student who merges "carbon forms many bonds" with "carbon forms many compounds" picks this. The maximum is four bonds per atom; the diversity comes from carbon atoms linking to each other repeatedly in chains and rings.
3. It transfers its four outer electrons to other atoms, giving charged compounds — A student who models every bond on ionic bonding picks this. Carbon forms covalent bonds, sharing a pair of electrons with each partner atom, so its compounds are uncharged molecules.
4. It links to neighbouring carbon atoms by weak hydrogen bonds that are easily broken — A student who thinks of hydrogen bonds as the general bond of biology picks this. Carbon atoms are joined by strong covalent bonds; hydrogen bonds are weak attractions between molecules or parts of molecules.

Syllabus statement B1.1.1 · Read this in Learn

1. A molecule of water is taken in and split to supply the atoms that join the monomers — A student who has the two reactions swapped picks this. Water is used and split in hydrolysis, which breaks bonds; condensation forms a bond and releases water.
2. A hydrogen bond forms between the two monomers, holding them together as a disaccharide — A student who thinks of hydrogen bonds as the bond of biology picks this. The bond formed by condensation is a strong covalent bond, which is why the disaccharide is stable until it is hydrolysed.
3. Water dissolves the two monomers so that they meet, but no water is formed or used — A student who sees water only as the solvent picks this. Water is a product of condensation: one molecule is released for every bond formed.
4. A covalent bond forms between the monomers and a molecule of water is released — In a condensation reaction an -OH from one monomer and an -H from the other are removed and combine to form water, leaving a covalent (glycosidic) bond between the monomers. Repeated condensations build a polysaccharide.

Syllabus statement B1.1.2 · Read this in Learn

1. Water acts only as the solvent in which enzymes break the polymer into its monomers — A student who sees water as the inert medium picks this. Water is a reactant that is split and incorporated into the products, which is exactly what the name hydrolysis records.
2. Water molecules are split, and the -H and -OH produced are added to the monomers — Hydrolysis means splitting with water. For each bond broken, one water molecule is split; its -H is added to one product and its -OH to the other, so the water is incorporated into the monomers.
3. Breaking the bonds between the monomers releases the energy that was stored in those bonds — A student who believes bond-breaking releases energy picks this. Hydrolysis releases very little energy; the energy from food is released later, when monomers are oxidized in cell respiration.
4. Water molecules are released as each bond between neighbouring monomers is broken — A student who has swapped the two reactions picks this. Water is released in condensation, when bonds form; in hydrolysis water is used up as bonds break.

Syllabus statement B1.1.3 · Read this in Learn

1. Both P and Q are pentose sugars — A student who counts the atoms in the ring picks this: both rings have five atoms. But one ring atom is oxygen and carbon atoms also lie outside the ring, so the ring size does not give the number of carbons.
2. Both P and Q are hexose sugars — A student who assumes every monosaccharide has six carbons, like glucose, picks this. P has only five carbon atoms in total, so it is a pentose.
3. P is a pentose and Q is a hexose — Pentoses and hexoses are named for their total number of carbon atoms. P has 4 + 1 = 5 carbon atoms, so it is a pentose (such as ribose); Q has 4 + 2 = 6 carbon atoms, so it is a hexose (such as fructose), even though its ring has only five atoms.
4. P and Q are alpha and beta glucose — A student who thinks any two differently drawn ring sugars are the alpha and beta forms picks this. Alpha- and beta-glucose have the same atoms and differ only in the orientation of one -OH group; P and Q differ in their number of carbon atoms.

Syllabus statement B1.1.4 · Read this in Learn

1. The glycosidic bonds within cellulose are very much stronger than those within starch — A student who looks for strength in stronger covalent bonds picks this. The glycosidic bonds in cellulose and starch are both covalent and of similar strength; the difference lies in how the chains are arranged and cross-linked.
2. The chains coil into tight helices that stretch and recoil like a spring — A student who transfers the coiled shape of amylose to cellulose picks this. Cellulose chains are straight, and it is their side-by-side bundling, not coiling, that gives strength.
3. Beta-glucose is a larger monomer than alpha-glucose, giving thicker chains — A student who thinks the two forms of glucose differ in size or atoms picks this. Alpha- and beta-glucose have identical formulae; beta-glucose differs only in the orientation of the -OH on carbon 1.
4. Straight chains lie side by side in bundles, cross-linked by many hydrogen bonds — The straight, unbranched chains of beta-glucose pack side by side and are grouped into bundles in which a very large number of hydrogen bonds link the chains. Although each hydrogen bond is weak, together they give the bundles great resistance to stretching.

Syllabus statement B1.1.6 · Read this in Learn

1. The A and B antigens differ only in the sugar at the end of their carbohydrate chain — ABO antigens are glycoproteins with a shared base carbohydrate chain. The alleles of the ABO gene encode enzymes that add different terminal monosaccharides to that chain, producing the A or the B antigen, and it is this small difference in the carbohydrate that other cells and antibodies recognize.
2. The A and B antigens are two entirely different proteins with unrelated amino acid sequences — A student who thinks of antigens only as proteins picks this. The identity of an ABO antigen lies in its carbohydrate chain, not its protein; A and B share the same base and differ in one terminal sugar.
3. Group O red blood cells carry a third antigen, the O antigen, in the place of the A or B — A student who extends the A/B naming pattern to O picks this. Group O cells carry the unmodified base chain and neither the A nor the B antigen; there is no O antigen.
4. The carbohydrate part of each antigen projects inwards into the cell's cytoplasm — A student who has not registered the orientation of glycoproteins picks this. The carbohydrate faces outwards from the cell surface, which is what allows it to act in cell–cell recognition and to be met by antibodies.

Syllabus statement B1.1.7 · Read this in Learn

1. Its molecules are too large to fit between water molecules, but a non-polar solvent has more room — A student who reuses the size explanation for polysaccharides picks this. A triglyceride is far smaller than a soluble protein; its insolubility in water comes from its non-polar nature, not its size.
2. Its molecules are mostly non-polar C–H bonds, with few groups that hydrogen-bond to water — Lipids are hydrophobic because they are non-polar. Water molecules hydrogen-bond to one another and exclude molecules that cannot join in, so a triglyceride is sparingly soluble in water; in a non-polar solvent, whose molecules interact by the same weak attractions, it dissolves readily.
3. Its molecules are less dense than water, so they float on the surface rather than dissolving — A student who merges floating with not mixing picks this. Density decides whether a liquid floats, not whether it dissolves: ethanol is less dense than water yet mixes with it completely.
4. It is a polymer of fatty acids, and polymers such as starch do not dissolve in water either — A student who classes triglycerides as polymers picks this. A triglyceride is one glycerol linked to three fatty acids, not a polymer, and its behaviour in solvents is explained by polarity, not by polymer size.

Syllabus statement B1.1.8 · Read this in Learn

1. Z, because its C=C bonds kink the chain so that molecules pack less closely together — Each C=C double bond puts a kink in the hydrocarbon chain. The kinked molecules of Z cannot pack closely, so the intermolecular attractions between them are weakest and less energy is needed to separate them: Z, the polyunsaturated fatty acid, melts at the lowest temperature.
2. X, because C=C bonds are stronger than single bonds and hold Y and Z molecules together — A student who confuses bonds within molecules with attractions between molecules picks this. C=C bonds are within each molecule; what matters for melting is how closely the molecules pack, and straight saturated chains such as X pack most closely and melt highest.
3. Z, because it has the fewest hydrogen atoms and so the fewest bonds to break as it melts — A student who counts the bonds to be broken picks this. Melting breaks no covalent bonds; it overcomes the weak attractions between molecules. Z does melt lowest, but only because its kinked chains pack least closely, not because it contains fewer bonds.
4. X, because having no double bonds makes it the unsaturated one, and unsaturated fats melt lowest — A student who takes "saturated" to mean full of double bonds picks this. A fatty acid is saturated with hydrogen, not with double bonds: X, with no C=C, is the saturated one, and its straight chains pack most closely, so it melts highest, not lowest.

Syllabus statement B1.1.10 · Read this in Learn

1. They release less energy per gram than carbohydrates, so a store of them lasts longer — A student who thinks of carbohydrate as "the energy food" picks this. Triglycerides release about twice as much energy per gram as carbohydrates; a store lasts because it is large and compact, not because it is low in energy.
2. They are very large molecules, so they cannot dissolve and leave the adipose cell — A student who explains lipid insolubility by size picks this. Triglycerides are insoluble because they are non-polar, not because they are large; a triglyceride is far smaller than a glycogen molecule.
3. They release about twice the energy per gram of carbohydrates when oxidized — Triglycerides yield roughly twice the energy per gram of carbohydrate, and they are stored without the water that accompanies glycogen and without osmotic effects, so a great deal of energy is held in a small, stable mass, which is what a long-term store needs.
4. Their many ester bonds store energy that is released when the bonds are broken — A student who believes energy is stored in bonds picks this. Breaking ester bonds by hydrolysis releases very little energy; the large yield comes from oxidizing the fatty acids and glycerol in cell respiration.

Syllabus statement B1.1.11 · Read this in Learn

1. Bilayers form, because the hydrophilic heads face the water and the hydrophobic tails face each other — Phospholipids are amphipathic. In water the hydrophilic phosphate heads associate with water while the hydrophobic fatty acid tails are excluded from it, so the molecules spontaneously arrange themselves in a double layer with heads outwards and tails inwards; no enzymes are needed.
2. Nothing organized forms, because phospholipid bilayers can only be built by enzymes inside living cells — A student who assumes every biological structure is built by enzymes picks this. Bilayer formation is a consequence of the hydrophobic and hydrophilic regions of the molecule and happens spontaneously in water.
3. Bilayers form, because covalent bonds link the fatty acid tails of neighbouring phospholipids — A student who expects an orderly structure to be covalently bonded picks this. No covalent bonds form between phospholipids; the bilayer is held by the exclusion of the tails from water, which leaves it fluid.
4. The phospholipids react with the water, because being amphipathic means acting as both an acid and a base — A student who confuses amphipathic with amphoteric picks this. Amphipathic means having a hydrophilic and a hydrophobic region, and it is this that drives bilayer formation; no acid–base reaction with water occurs.

Syllabus statement B1.1.12 · Read this in Learn

1. 70 nm, because each prefix is ten times the next — A student who treats each SI prefix as a factor of ten picks this. Micro is 10⁻⁶ and nano is 10⁻⁹, so the two differ by a factor of 1000, not 10.
2. 7000 nm, because one micrometre is 1000 nanometres — 1 µm = 10⁻⁶ m and 1 nm = 10⁻⁹ m, so 1 µm = 1000 nm. The length is unchanged; it is expressed in a smaller unit, so the number is larger: 7 × 1000 = 7000 nm.
3. 0.007 nm, because a smaller unit means a smaller number — A student who divides because "nano is smaller" picks this. A smaller unit fits into the same length more times, so converting micrometres to nanometres multiplies by 1000.
4. 7 000 000 nm, because a micrometre is 10⁻³ of a metre — A student who has swapped milli and micro picks this, treating 7 µm as 7 × 10⁻³ m. Micro is 10⁻⁶, so 7 µm is 7 × 10⁻⁶ m, which is 7000 nm.

Syllabus statement B1.1.1 · Read this in Learn

1. Their meanings were discovered by scientists measuring natural objects — A student who treats every part of science as discovered fact picks this. The prefixes are conventions that could have been defined differently; what is discovered is the size of the objects measured with them.
2. Each prefix is one tenth of the prefix that comes before it — A student who generalizes from centi and milli, which really are a factor of ten apart, picks this. Milli (10⁻³), micro (10⁻⁶) and nano (10⁻⁹) are each a factor of 1000 apart.
3. Their meanings are conventions fixed by international agreement — The values of the SI prefixes are scientific conventions agreed internationally so that a measurement means the same thing to every scientist. They are agreed, not discovered, which is a nature-of-science point the guide makes explicitly.
4. "micro" means one thousandth, and "milli" means one millionth — A student who has the two m-prefixes swapped picks this. Milli is one thousandth (10⁻³) and micro is one millionth (10⁻⁶); that is why cells are measured in micrometres.

Syllabus statement B1.1.1 · Read this in Learn

1. 99 water molecules — A chain of 100 amino acids contains 99 peptide bonds. Each hydrolysis reaction breaks one bond and uses one water molecule, which is split to supply the -H and -OH added to the products, so 99 water molecules are used.
2. 100 water molecules — A student who counts one bond per monomer picks this. Bonds lie between monomers, so 100 monomers are joined by 99 bonds, and hydrolysing them uses 99 water molecules.
3. 1 water molecule — A student who reads "polymer + water → monomers" as a single reaction picks this. Each bond needs its own hydrolysis reaction and its own water molecule.
4. 0 water molecules — A student who thinks water is merely the solvent picks this. Water is a reactant in hydrolysis: one molecule is consumed and split for every bond broken.

Syllabus statement B1.1.3 · Read this in Learn

1. Glucose breaks down readily on its own, which is why it is used as a respiratory substrate — A student who equates energy-rich with unstable picks this. Glucose is chemically stable; its energy is released only by controlled, enzyme-catalysed oxidation in cell respiration.
2. Glucose dissolves in water because its molecules are small, so it is easily transported — A student who uses size as the rule for solubility picks this. Glucose dissolves because its polar hydroxyl groups form hydrogen bonds with water; small non-polar molecules such as steroids do not dissolve.
3. Glucose is highly soluble, so cells store large quantities of it dissolved in their cytoplasm — A student who does not consider osmotic effects picks this. Cells convert glucose into relatively insoluble polysaccharides for storage; a high concentration of dissolved glucose would draw in water by osmosis.
4. Glucose is chemically stable, so it is transported in blood plasma without breaking down — Glucose does not break down spontaneously; its covalent bonds are broken only when enzymes act on it in cell respiration. This stability, together with its solubility, is what allows it to be carried in the blood to the cells that oxidize it.

Syllabus statement B1.1.4 · Read this in Learn

1. Glycogen is relatively insoluble because of its size, so it does not affect the osmotic balance of the cell — Glycogen is a very large molecule and is relatively insoluble, so large amounts can be stored without raising the solute concentration of the cytoplasm. Free glucose is soluble and would draw water into the cell by osmosis.
2. Glycogen is insoluble because it is non-polar, so it stays separate from the watery cytoplasm — A student who reuses the lipid explanation for insolubility picks this. Glycogen is covered in polar hydroxyl groups; it is relatively insoluble only because its molecules are so large.
3. The branches of glycogen make its glucose hard to remove, so the store is not used up too quickly — A student who imagines branches as protection picks this. Branching does the opposite: it provides many chain ends at which glucose can be removed rapidly by hydrolysis.
4. Glycogen stores extra energy in the glycosidic bonds that link its glucose monomers together — A student who believes energy is stored in bonds picks this. Polymerizing glucose does not add energy; the energy is released when glucose is oxidized in cell respiration, and glycogen is simply a compact, osmotically inert way of holding it.

Syllabus statement B1.1.5 · Read this in Learn

1. Branches shield the glycosidic bonds, slowing hydrolysis so that the store is released steadily over time — A student who sees branches as protection picks this. Branches expose more chain ends, speeding hydrolysis; a store that was released slowly would not suit an animal needing glucose quickly.
2. Monomers are removed by hydrolysis at chain ends, so more branches give more ends to act on at once — Glucose is removed from a polysaccharide only at the ends of chains. Every branch point creates an extra end, so a molecule with a branch every 10 monomers can be hydrolysed at many more points simultaneously than amylopectin, matching an animal's need for rapid mobilization.
3. Branching makes glycogen non-polar, so that it remains insoluble until the glucose is actually required — A student who explains insolubility by non-polarity picks this. Glycogen is a polar molecule covered in hydroxyl groups; it is relatively insoluble because of its size, and branching has nothing to do with polarity.
4. Each branch point adds a bond in which extra energy is stored for release when the bond is broken — A student who believes bonds store energy picks this. Breaking glycosidic bonds absorbs a little energy rather than releasing it; the energy comes from oxidizing the glucose released, however many branches there are.

Syllabus statement B1.1.5 · Read this in Learn

1. Beta-glucose has a different molecular formula, and its extra atoms make the chain rigid — A student who believes alpha- and beta-glucose are different sugars picks this. Both are C₆H₁₂O₆; they differ only in the orientation of one -OH group, and that alone produces the alternating, straight chain.
2. Covalent bonds between neighbouring cellulose chains pull each chain out straight and flat — A student who expects covalent cross-links picks this. Neighbouring chains are linked only by hydrogen bonds, and the straightness of each chain comes from the alternating orientation of its own monomers.
3. Each beta-glucose is rotated 180° relative to its neighbours, so the chain is straight — In beta-glucose the -OH on carbon 1 points above the ring, so successive monomers must alternate in orientation (each rotated 180°) to form the glycosidic bond. The alternating orientation cancels any tendency to curve, giving straight chains that can be bundled and cross-linked by hydrogen bonds.
4. Branches along the cellulose chain, as in amylopectin, prevent the chain from coiling — A student who transfers the structure of starch to cellulose picks this. Cellulose chains are unbranched; they are straight because of the alternating orientation of beta-glucose monomers.

Syllabus statement B1.1.6 · Read this in Learn

1. One glycerol, three fatty acids and one phosphate group — A student who reads "phosphate added to a triglyceride" literally picks this. Glycerol can link only three components, so the phosphate replaces one fatty acid rather than being added to three.
2. Three glycerol molecules, one fatty acid and one phosphate group — A student who reads the tri- of triglyceride as three glycerols picks this pattern. Lipids built on glycerol have one glycerol; it is the number of fatty acids that varies.
3. One glycerol, two fatty acids and one phosphate group — Glycerol has three hydroxyl groups. In a phospholipid two of them are linked by condensation to fatty acids and the third to a phosphate group, giving a molecule with two hydrophobic tails and a hydrophilic head.
4. Many fatty acid monomers joined in a chain, plus one phosphate group — A student who assumes lipids are polymers picks this. A phospholipid is not a chain of monomers: it has a fixed structure of one glycerol, two fatty acids and one phosphate group.

Syllabus statement B1.1.9 · Read this in Learn

1. One water molecule is released — A student who imagines the whole molecule forming in a single reaction picks this. Three separate bonds form, and each condensation releases its own water molecule.
2. Four water molecules are released — A student who counts one water molecule per component picks this. Water is released per bond formed, and four components are joined by three bonds.
3. No water is released; water is used up — A student who has swapped condensation and hydrolysis picks this. Condensation releases water as bonds form; it is hydrolysis, the reverse reaction, that uses water up.
4. Three water molecules are released — Each of the three hydroxyl groups of glycerol reacts with the carboxyl group of one fatty acid in a separate condensation reaction, forming an ester bond and releasing one water molecule, so three water molecules are released in total.

Syllabus statement B1.1.9 · Read this in Learn

1. The mammal's fat must be solid inside its body, so it requires the higher melting point that a high proportion of saturated fatty acids provides, whereas the seed's oil only needs to be liquid at room temperature — A student who judges the state of a stored fat at room temperature picks this. The relevant temperature is the mammal's body temperature, at which its fat is sufficiently fluid; being solid at 20°C says nothing about its state at 37°C.
2. The mammal keeps a high, constant body temperature at which a store rich in saturated fatty acids stays fluid enough, while the seed is at lower environmental temperatures and needs unsaturated fatty acids with low melting points — Saturated fatty acids raise the melting point; unsaturated ones lower it. An endotherm maintains a body temperature well above 20°C, so a store with a high proportion of saturated fatty acids remains sufficiently fluid inside it. A plant is at environmental temperature, which is often low, so its store must be an oil rich in unsaturated fatty acids that stays liquid.
3. The mammal's fat is solid because saturated fatty acids contain many C=C double bonds, which raise the melting point, whereas the unsaturated fatty acids of the oil have none and so remain liquid — A student who thinks "saturated" means full of double bonds picks this. Saturated fatty acids have no C=C bonds and pack closely, which is why they raise the melting point; it is the unsaturated fatty acids that have C=C bonds.
4. The oil is liquid because its unsaturated fatty acids have fewer hydrogen atoms and so fewer covalent bonds to break when they melt, whereas the mammal's saturated fatty acids have more bonds to break and stay solid — A student who thinks melting breaks covalent bonds picks this. Melting leaves every molecule intact and only overcomes the attractions between molecules; the oil melts lower because the kinks from its C=C bonds stop the chains packing closely.

Syllabus statement B1.1.10 · Read this in Learn

1. All animals in cold water need thick insulation, so the cod must have an equally thick layer of fat elsewhere in its body to keep itself warm in the polar sea — A student who links insulation to habitat alone picks this. The cod is an ectotherm with a body temperature close to the water's, so a fat layer could not keep it warm; only endotherms, whose bodies are warmer than their habitat, benefit from thick insulation.
2. Blubber generates the heat that keeps the seal warm in polar water, the dolphin needs to generate less heat in warm water, and the cod does not generate any heat at all — A student who thinks fat produces warmth picks this. Blubber does not generate heat; it slows the loss of heat produced by metabolism. Its thickness reflects how much heat loss must be prevented, not how much heat must be made.
3. Blubber traps a layer of air against the skin, the dolphin's warmer water means that less air is needed, and the cod's scales cannot hold a layer of air in place under water — A student who transfers the trapped-air mechanism of fur and feathers picks this. Blubber contains no air; the triglyceride-filled tissue is itself a poor conductor, which is why it insulates under water where trapped air would be lost.
4. Adipose tissue slows heat loss, so an endotherm needs more of it the larger the gap between its body and habitat temperatures, and the cod's body is at almost the temperature of the water — Blubber is a poor conductor of heat, so it reduces the rate of heat loss from a warm body to colder surroundings. The seal maintains a body temperature far above that of polar water and needs thick insulation; the dolphin in warm water loses heat more slowly and needs less; the cod is an ectotherm at almost the temperature of the water, so insulation would retain no heat.

Syllabus statement B1.1.11 · Read this in Learn

1. It is too large to pass between the phospholipids, so it is carried through a channel protein — A student who applies the rule that only small molecules cross the bilayer picks this. Polarity matters more than size here: non-polar steroids dissolve in the bilayer itself and need no channel.
2. It is a non-polar steroid, so it dissolves in the hydrophobic core of the bilayer and diffuses across — Testosterone is a steroid, a lipid built on four fused carbon rings, and is non-polar. It dissolves in the hydrophobic interior of the phospholipid bilayer and passes through by simple diffusion, which is why steroid hormones can act on receptors inside the cell.
3. It is a protein hormone, so it is pulled into the cell after binding to a receptor on the surface — A student who assumes all hormones are proteins picks this. Testosterone is a steroid lipid, not a protein; protein hormones cannot cross the bilayer, whereas steroids can.
4. It is enclosed in a vesicle at the cell surface and taken into the cell by endocytosis — A student who defaults to endocytosis for anything entering a cell picks this. A steroid needs no vesicle: being non-polar it diffuses straight through the hydrophobic core of the membrane.

Syllabus statement B1.1.13 · Read this in Learn

1. A single six-membered ring made up of five carbon atoms and one oxygen atom — A student who treats any ring as the mark of a steroid picks this. A single ring containing an oxygen atom, with carbon outside it, is the ring form of a hexose such as glucose.
2. A chain of amino acids folded up into a compact globular shape — A student who thinks steroid hormones are proteins picks this. A folded amino acid chain is a protein; steroids are lipids with four fused rings and no amino acids.
3. Four fused rings of carbon atoms, three with six atoms and one with five — A steroid is recognized by its four fused rings sharing edges, three six-membered and one five-membered. Oestradiol and testosterone both have this structure, which is non-polar and lets them pass through the phospholipid bilayer.
4. A glycerol backbone joined to three long hydrocarbon chains — A student who equates lipids with fats picks this, choosing the fat. This is a triglyceride; steroids are a different class of lipid with a fused-ring structure rather than fatty acid chains.

Syllabus statement B1.1.13 · Read this in Learn